On Mon, Jul 06, 2020 at 09:36:25AM +0200, Olivier Matz wrote:
> Hi Hongzhi,
>
> I suggest the following title instead:
>
> net: fix checksum on big endian CPUs
>
> On Wed, Jun 24, 2020 at 05:11:19PM +0200, Morten Brørup wrote:
> > > From: dev [mailto:dev-boun...@dpdk.org] On Behalf Of Thomas Monjalon
> > > Sent: Wednesday, June 24, 2020 5:04 PM
> > >
> > > 24/06/2020 15:00, Morten Brørup:
> > > > > From: Thomas Monjalon [mailto:tho...@monjalon.net]
> > > > > Sent: Wednesday, June 24, 2020 2:22 PM
> > > > >
> > > > > 27/05/2020 15:40, guohongzhi:
> > > > > > From: Hongzhi Guo <guohongz...@huawei.com>
> > > > > >
> > > > > > __rte_raw_cksum should consider Big Endian.
> > > > >
> > > > > We need to explain the logic in the commit log.
>
> Here is a suggestion of commit log:
>
> With current code, the checksum of odd-length buffers is wrong on
> big endian CPUs: the last byte is not properly summed to the
> accumulator.
>
> Fix this by left-shifting the remaining byte by 8. For instance,
> if the last byte is 0x42, we should add 0x4200 to the accumulator
> on big endian CPUs.
>
> This change is similar to what is suggested in Errata 3133 of
> RFC 1071.
>
> Can you please submit a new version with the 2 changes above?
One more thing, please also add:
Fixes: 6006818cfb26 ("net: new checksum functions")
Cc: sta...@dpdk.org
Thanks
Olivier
>
> > > >
> > > > Having grown up with big endian CPUs, reading the final byte like
> > > this is obvious to me. I struggle understanding the little endian way
> > > of reading the last byte. (Not really anymore, but back when little
> > > endian was unfamiliar to me I would have struggled.)
> > > >
> > > > An RFC (I can't remember which) describes why the same checksum
> > > calculation code works on both big and little endian CPUs. Is it this
> > > explanation you are asking for?
> > >
> > > This explanation may be interesting.
> > >
> >
> > RFC 1071, especially chapter 3.
> >
> > Please note that big endian is considered "Normal" order in the RFC. :-)
>
> There is an errata for this RFC about the C code:
> see https://www.rfc-editor.org/errata/eid3133
>
> > >
> > > > > > Signed-off-by: Hongzhi Guo <guohongz...@huawei.com>
> > > > > > ---
> > > > > > +#if (RTE_BYTE_ORDER == RTE_BIG_ENDIAN)
> > > > > > + sum += *((const uint8_t *)u16_buf) << 8;
> > > > > > +#else
> > > > > > sum += *((const uint8_t *)u16_buf);
> > > > > > +#endif
> > > > >
> > > > > *((const uint8_t *)u16_buf) should be an uint8_t.
> > > > > What is the expected behaviour of shifting 8 bits of a byte?
> > > >
> > > > Yes, the value will be an uint8_t type. But the shift operation will
> > > cause the compiler to promote the type to int before shifting it.
> > >
> > > This is the explanation I was looking for :-)
> > >
> > >
> >
